∫(0，π)sin²xdx

=∫(0，π)[1-cos(2x)]/2

dx

=∫(0，π)[1-cos(2x)]/4

d(2x)

=(1/4)∫(0，π)[1-cos(2x)]d(2x)

=(1/4)

[2x-sin(2x)/2]

|(0，π)

=(1/4)[2π-sin(2π)/2-2×0-sin(0)/2]

=(1/4)(2π)

=π/2

提示：利用三角函数公式cos(2x)=1-2sin²x，再将d(x)换成(1/2)d(2x)，剩下的就好办了。